JEE Main Electromagnetic Waves Previous Year Questions with Solutions

Electromagnetic waves or EM waves have a vibrating electric field and magnetic field. These are the waves that propagate at the speed of light in empty space. The speed of light is equal to 3 x 108 m/s. Some of the electromagnetic waves like visible light are harmless.

X-ray is an example of an EM wave that can be harmful if the person is exposed to it for a longer time.

JEE Main solved questions on Electromagnetic Waves

Electromagnetic waves are represented as a sinusoidal graph. The electric and magnetic fields are perpendicular to each other and are also perpendicular to the direction of propagation of waves. Electromagnetic waves are transverse in nature. The highest point of the wave is known as crest while the lowest point is known as the trough.

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JEE Main Previous Year Solved Questions on Electromagnetic Waves

Q1: The mean intensity of radiation on the surface of the Sun is about 108 W/m2. The RMS value of the corresponding magnetic field is closest to

(a) 10–2 T

(b) 1 T

(c) 10–4 T

(d) 102 T

Solution

Mean intensity = (B2rms0)c

B2rms = (108 x 4π x 10-7)/(3 x 108)

Brms = 10-4 T

Answer: (c) 10–4 T

Q2: A plane electromagnetic wave travels in free space along the x-direction. The electric field component of the wave at a particular point of space and time is E = 6 V m–1 along the y-direction. Its corresponding magnetic field component, B would be

(a) 6 × 10–8 T along the x-direction

(b) 2 × 10–8 T along the y-direction

(c) 2 × 10–8 T along the z-direction

(d) 6 × 10–8 T along the z-direction

Solution

The direction of electromagnetic wave travelling is given by

kˉ=Eˉ×Bˉ\bar{k} = \bar{E} \times \bar{B}

As the wave is travelling along x-direction and E is along the y-direction.

So B must point towards the z-direction.

The magnetic field, B = E/c = 6/(3 x 108) = 2 x 10-8 T

Answer: (b) 2 × 10–8 T along the y-direction

Q3: 50 W/m2 energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy (25%) is reflected from the surface and the rest is absorbed. The force exerted on 1m2 surface area will be close to (c = 3 × 108 m/s)

(a) 20 × 10–8 N

(b) 10 × 10–8 N

(c) 35 × 10–8 N

(d) 15 × 10–8 N

Solution

Given energy density = 50 W/m2

Here, change in momentum Δp = pf – pi

Δp = (-pi /4) – pi

Δp = -5pi /4 ∵ pi = E/c = 50 W/s/(3 x 108)

Δp/Δt = F = 丨-5pi/4丨= (5/4) x (50/(3 x 108) = 20.8 x 10-8 N ≈ 20x 10-8 N

Answer: (a) 20 × 10–8 N

Q4: A red LED emits light at 0.1 watts uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is

(a) 5.48 V/m

(b) 7.75 V/m

(c) 1.73 V/m

(d) 2.45 V/m

Solution

The intensity of light, I = uav c

Also, I = P/4πr2 and uav = ½ε0E02

∴ P/4πr2 = (½)ε0E02

Or E0=2P4πϵ0r2cE_{0} =\sqrt{\frac{2P}{4\pi \epsilon _{0}r^{2}c}}

Here, P = 0.1 W, r = 1 m, c = 3 × 108 m s–1

1/4πε0 = 9 x 109 N C–2 m2

E0=2×0.1×9×10912×3×108=6E_{0} =\sqrt{\frac{2\times 0.1 \times 9\times 10^{9}}{1^{2}\times 3\times 10^{8}}} = \sqrt{6}

= 2.45 V m-1

Answer: (d) 2.45 V/m

Q5: Match List-I (Electromagnetic wave type) with List-II (Its association/application) and select the correct option from the choices given below the lists

List-I List-II
(P) Infrared waves (i) To treat muscular strain
(Q) Radio waves (ii) For broadcasting
(R) X-rays (iii) To detect fracture of bones
(S) Ultraviolet rays (iv) Absorbed by the ozone layer of the atmosphere

P   Q    R     S

(a) (i) (ii) (iii) (iv)

(b) (iv) (iii) (ii) (i)

(c) (i) (ii) (iv) (iii)

(d) (iii) (ii) (i) (iv)

Solution

Infrared waves are used to treat muscular strain. Radio waves are used for broadcasting. X-rays are used to detect the fracture of bones. Ultraviolet rays are absorbed by the ozone layer of the atmosphere.

Answer: (a)

P   Q     R      S

(i) (ii) (iii) (iv)

Q6: During the propagation of electromagnetic waves in a medium

(a) both electric and magnetic energy densities are zero

(b) electric energy density is double of the magnetic energy density

(c) electric energy density is half of the magnetic energy density

(d) electric energy density is equal to the magnetic energy density

Solution: In an em wave, energy is equally divided between the electric and the magnetic fields.

Answer: (d) The electric energy density is equal to the magnetic energy density

Q7: The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is

(a)12 V/m

(b) 3 V/m

(c) 6 V/m

(d) 9 V/m

Solution

In an electromagnetic wave, the peak value of the electric field (E0) and peak value of magnetic field (B0) are related by

E0 = B0C

E0 = (20 × 10–9 T) (3 × 108 m s–1) = 6 V/m

Answer: (c) 6 V/m

Q8: An electromagnetic wave of frequency = 3.0 MHz passes from vacuum into a dielectric medium with permittivity = 4.0. Then

(a) wavelength is doubled and the frequency remains unchanged

(b) wavelength is doubled and frequency becomes half

(c) wavelength is halved and frequency remains unchanged

(d) wavelength and frequency both remain unchanged

Solution

During propagation of a wave from one medium to another, the frequency remains constant and wavelength changes.

μ=ϵϵ0=4\mu =\sqrt{\frac{\epsilon }{\epsilon _{0}}}=\sqrt{4} = 2

Since μ = 1/λ

Wavelength is halved

Answer: (c) wavelength is halved and frequency remains unchanged

Q9: Electromagnetic waves are transverse in nature is evident by

(a) polarization

(b) interference

(c) reflection

(d) diffraction

Answer: (a) Polarization proves the transverse nature of electromagnetic waves.

Q10: The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then

(a) UE > UB

(b) UE = UB/2

(c) UE = UB

(d) UE < UB

Answer: (c) UE = UB

Q11: A 27 mW laser beam has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by [Given permittivity of space ε0 = 9 × 10–12 SI units, speed of light c = 3 × 108 m/s]

(a) 2 kV/m

(b) 0.7 kV/m

(c) 1 kV/m

(d) 1.4 kV/m

Solution

The intensity of the electromagnetic wave is given by

I = Power(P)/Area(A) = ½ ε0 E2C

E=2PAϵ0c=2×27×103105×9×3×108E = \sqrt{\frac{2P}{A\epsilon _{0}c}} = \sqrt{\frac{2 \times 27 \times 10^{-3}}{10^{-5}\times 9\times 3\times 10^{8}}}

E = 1.4 kV/m

Answer: (d) 1.4 kV/m

Q12: The RMS value of the electric field of the light coming from the sun is 720 N/C. The average total energy density of the electromagnetic wave is

(a) 3.3 × 10–3 J/m3

(b) 4.58 × 10–6 J/m3

(c) 6.37 × 10–9 J/m3

(d) 81.35 ×10–12 J/m3

Solution

U = (½)є0 (E2rms)+ (½μ0) (B2rms)

U = (½)є0 (E2rms) + (½μ0) (E2rms/c2)

U = (½)є0 (E2rms) + (½μ0) (E2rmsє0μ0)

U = (½)є0 (E2rms) + (½)є0 (E2rms) = є0 (E2rms)

U = (8.85 x 10-12) x (720)2 = 4.58 x10-6 Jm-1

Answer (b) 4.58 × 10–6 J/m3

Q13: Which of the following are not electromagnetic waves?

(a) cosmic rays

(b) gamma rays

(c) β-rays

(d) X-rays

Answer: (c) β-rays are not electromagnetic waves

Q14: If a source of power 4kW produces 1020 photons/second, the radiation belongs to a part of the spectrum called

(a) microwaves

(b) y-rays

(c) X-rays

(d) ultraviolet rays

Solution

Eproton = (4 x 103)/1020 = 4 x 10-17 J = (4/1.6) x 102 eV = 250 eV

λproton = 1242/250 = 50 Å (x -ray)

Answer: (c) X-rays

Q15: An EM wave from the air enters a medium. The electric fields are

E1=E01x^cos[2πf(zct)]\vec{E_{1}} = E_{01}\hat{x}cos\left [ 2\pi f(\frac{z}{c}-t) \right ]

in air and E2=E02x^cos[k(2zct)]\vec{E_{2}} = E_{02}\hat{x}cos\left [ k(2z-ct) \right ]

in medium, where the wavenumber k and frequency f refer to their values in air. The medium is non-magnetic. If εr1 and εr2 refer to relative permittivities of air and medium respectively, which of the following options is correct?

(a) εr1r2 = 4

(b) εr1r2 = 2

(c) εr1r2 = 1/4

(d) εr1r2 = ½

Solution

In the air, the EM wave is

E1=E01x^cos[2πf(zct)]\vec{E_{1}} = E_{01}\hat{x}cos\left [ 2\pi f(\frac{z}{c}-t) \right ]

= E2=E02x^cos[k(zct)]\vec{E_{2}} = E_{02}\hat{x}cos\left [ k(z-ct) \right ] (since, k = 2π/λ0 = 2πf/c)

In the medium, the EM wave is

E2=E02x^cos[k(2zct)]\vec{E_{2}} = E_{02}\hat{x}cos\left [ k(2z-ct) \right ] E2=E02x^cos[2k(z(c/2)t)]\vec{E_{2}} = E_{02}\hat{x}cos\left [ 2k(z-(c/2)t) \right ]

During refraction, frequency remains unchanged, whereas the wavelength gets changed

k’ = 2k (From equations)

2π/λ’ = 2(2π/λ0) = λ’= λ0/2

Since, v = c/2

1μ0ϵr2=12×1μ0ϵr1\frac{1}{\sqrt{\mu _{0}\epsilon_{r2}} } = \frac{1}{2}\times \frac{1}{\sqrt{\mu _{0}\epsilon_{r1}} }

εr1r2= 1/4

Answer: (c) εr1r2 = 1/4